In the experiment set up of meter bridge shown in the figure. The null point is obtained at a distance of $$40$$ cm from A . If a $$12$$Ω resistor is connected in series with $$R1$$, the null point shifts by $$10$$ cm. The resistance that should be connected in parallel with $$R1+12$$Ω such that the null point shifts back to initial position is

Question

In the experiment set up of meter bridge shown in the figure. The null point is obtained at a distance of $$40$$ cm from A . If a $$12$$Ω resistor is connected in series with $$R1$$, the null point shifts by $$10$$ cm. The resistance that should be connected in parallel with $$R1+12$$Ω such that the null point shifts back to initial position is

Infinity Learn · Accepted Answer

Initially,  $$R1R2=l1l2$$ ⇒$$R1R2=(40)(100-40)=4060=23$$→$$1$$  Now , $$12$$ Ω is connected in series with $$R1$$ .  This increases the value of resistance on left side and null point shifts towards right by $$10$$ cm. The new balancing length becomes $$50$$ cm. ∴$$R1+12R2=(50)(100-50)$$ $$=$$ $$1$$ ⇒$$R1+12=R2$$→$$2$$  Using (1) and (2) ,  $$2R23+12=R2$$ ⇒$$12=R23$$⇒$$R2=36$$Ω  and from (1),    $$R1=24$$Ω   Finally,  let say "R" is connected in parallel with $$R1+12$$ $$=$$ $$24+12=36$$Ω. As a result, null point should shift leftwards by $$10$$ cm  making balancing length again $$40$$ cm  ∴ $$ReqR2=(40)(100-40)$$ ⇒$$36$$×$$R36+R36=23$$ ⇒$$R=72$$Ω

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